The first one is categorical data
the second one is discrete numerical
and the third one is continuous numerical
i hope this helps you
<span />
Answer:
2
Step-by-step explanation:
Given the question :
Serena wants to create snack bags for a trip she is going on. She has 6 granola bars and 10 pieces of dried fruit. If the snack bags should be identified without any food leftover, what is the greatest number of snack bags Serena can make?
Number of granolas = 6
Number of dried fruits = 10
Since the snackbag is to be designed in such a way that there should be no food leftover, the greatest number of snack bags Serena can make could be obtained by getting the highest common factor of (6 and 10)
____6____10
2___3____5
Here, the highest common factor of 6 and 10 is 2
Hence, the greatest number of snack bags she can make is 2.
Yeah... I'm normal good at angles, but the way you've taken the picture, it's pretty hard for us to figure out. All you need to do is grab you protractor and start measuring. Pretty simple.<span />
Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: number of daily text messages a high school girl sends.
This variable has a population standard deviation of 20 text messages.
A sample of 50 high school girls is taken.
The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:
X[bar]≈N(μ;δ²/n)
This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:
Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)
a.
P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836
b.
P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)
P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328
I hope you have a SUPER day!
